Problem solving?!

Question

In a certain course, there are two exams worth 30% each, a major paper worth 25% and quizzes worth 15%. If a student got scores of 84 on the first exam, and a 78 on the paper, and an average of 72 on the quizzes, what is the lowest score the student can get on the second test to have a semester average of 82?

Answer 1

.3*84+.3x+.25*78+.15*72   =  .82

30+30+25+15

 

.3*84+.3x+.25*78+.15*72= 82

      

55.5 + .3x = 82

 

.3x = 26.5

 

x ≈ 88.333

 

Answer 2

Brenda,

 

Let E₂ = minimum score needed on 2nd Exam to achieve desired overall weighted average, W

     W  = total weighted average = sum of pairs (weighting in decimal X score) for each test, paper & quiz.

 

So  W = 0.30E₂ + 0.30(84) + 0.25(78) + 0.15(72)  = 82

Then      0.30E₂ = 82 - 25.2 - 19.5 - 10.8 where these are the desired W, and the weighted contributions from Exam1, the Paper, and the Quizes.

             0.32E₂  = 82 - 55.5 = 26.5

                    E₂  = 26.5/0.30  =  88.33

 

So the answer rounded up is you need a minimum score of 89 on the second Exam to have a final average of 82.